Discussion of a conic depending on a parameter

Text

Discuss, as $k$ varies, which conic is described: $(25 - k^2) (x-1)^{2} + (1 - k) (y+1)^{2} = (2 - k)(k+3).$ Distinguish: ellipse, hyperbola (giving the focal axis), pair of intersecting lines, pair of parallel lines, a point, or the empty set.

Solution

Let $A=25-k^2=(5-k)(5+k)$, $B=1-k$, $C=(2-k)(k+3)$. The centre is always $(1;-1)$. Signs change at $k=-5,-3,1,2,5$. With $A(x-1)^2+B(y+1)^2=C$:

• $k<-5$: $A<0,\,B>0,\,C<0$ → hyperbola, foci on the horizontal axis.
• $k=-5$: $A=0$, $B(y+1)^2=C<0$ → empty set.
• $-5<k<-3$: $A>0,\,B>0,\,C<0$ → empty set.
• $k=-3$: $C=0$, $A,B>0$ → a point $(1;-1)$.
• $-3<k<1$: $A,B,C>0$ → ellipse (foci on the vertical axis, since $A>B$).
• $k=1$: $B=0$ → pair of parallel vertical lines.
• $1<k<2$: $A>0,\,B<0,\,C>0$ → hyperbola, foci on the horizontal axis.
• $k=2$: $C=0$, $A>0,\,B<0$ → pair of intersecting lines.
• $2<k<5$: $A>0,\,B<0,\,C<0$ → hyperbola, foci on the vertical axis.
• $k=5$: $A=0$, $B(y+1)^2=C<0$ with $B<0$ → pair of parallel horizontal lines.
• $k>5$: $A,B,C<0$ → ellipse (foci on the horizontal axis); at $k=\tfrac{1+\sqrt{97}}{2}$ we get $A=B$, i.e. a circle.

$\boxed{\text{see the interval-by-interval discussion above}}$